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SpyIntel [72]
3 years ago
8

Please see my other questions

Chemistry
2 answers:
Musya8 [376]3 years ago
7 0

Answer:

Okay I will go look at it

sdas [7]3 years ago
6 0

Answer: alr, gimme a sec

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A sample of ethanol (C2H6O) has a mass of 0.2301 g. Complete combustion of this sample causes the temperature of a bomb calorime
Keith_Richards [23]

Answer:  4.994\times 10^{-3}  moles

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass , occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}

Given mass of ethanol = 0.2301

Molar mass of ethanol = 46.07 g/mol

\text{Number of moles of ethanol}=\frac{0.2301g}{46.07g/mol}=4.994\times 10^{-3}

Thus there are 4.994\times 10^{-3}  moles of ethanol are present in the sample.

7 0
4 years ago
Read 2 more answers
A measure of how closely individual measurements agree with one another 1 point<br> is
rewona [7]

Precision

Precision: A measure of how closely individual measurements agree with one another. Accuracy: Refers to how closely individual measurements agree with the correct or true value.

6 0
3 years ago
An intravenous infusion is to contain 15 mEq of potassium ion and 20 mEq of sodium ion in 500 mL of 5% dextrose injection. Using
Studentka2010 [4]

Answer:

To supply the required ions it is necessary to inject 5,6mL of 6g/30mL solution and 131,1 mL of 0,9% solution.

Explanation:

1mEq of sodium are 59mg of NaCl and 1mEq of potassium are 75mg KCl

in intravenous infusion 15 mEq of K are:

15x75mg KCl = 1,125g of KCl

And 20 mEq of Na are:

20x59mg NaCl = 1,18g of NaCl

To supply the potassium ion it is necessary to inject:

1,125g of KCl×\frac{30mL}{6g} =<em> 5,6mL  of 6g/30mL solution</em>

And, to supply the sodium ion it is necessary to inject:

1,18g of NaCl×\frac{100mL}{0,9g} = <em>131,1 mL of 0,9% solution</em>

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I hope it helps!

6 0
4 years ago
The activation energy for the isomerization ol cyclopropane to propene is 274 kJ/mol. By what factor does the rate of this react
Bond [772]

Answer:

The rate of the reaction increased by a factor of 1012.32

Explanation:

Applying Arrhenius equation

ln(k₂/k₁) = Ea/R(1/T₁ - 1/T₂)

where;

k₂/k₁ is the ratio of the rates which is the factor

Ea is the activation energy = 274 kJ/mol.

T₁ is the initial temperature = 231⁰C = 504 k

T₂ is the final temperature = 293⁰C = 566 k

R is gas constant = 8.314 J/Kmol

Substituting this values into the equation above;

ln(k₂/k₁) = 274000/8.314(1/504 - 1/566)

ln(k₂/k₁) = 32956.4589 (0.00198-0.00177)

ln(k₂/k₁)  = 6.92

k₂/k₁ = exp(6.92)

k₂/k₁ = 1012.32

The rate of the reaction increased by 1012.32

3 0
4 years ago
Read 2 more answers
Would the density of a person be the same on the earth and on the moon ? explain
rosijanka [135]

Answer:

No

Explanation:

The density of a person stays the same no matter where they are in the universe

6 0
4 years ago
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